Understanding (the wiki page on) Verdier duality - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:19:14Z http://mathoverflow.net/feeds/question/68647 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/68647/understanding-the-wiki-page-on-verdier-duality Understanding (the wiki page on) Verdier duality James D. Taylor 2011-06-23T19:48:20Z 2012-04-19T01:11:04Z <p>My familiarity with concepts related to derived categories is only tangential, and little by little I intend to get more comfortable with them. I was playing around with Caldararu's introduction to the topic, and looking up various things on the web.</p> <p>Here is my question (that I have every confidence is trivial for experts):</p> <p>On the wiki page on Verdier duality <a href="http://en.wikipedia.org/wiki/Verdier_duality" rel="nofollow">http://en.wikipedia.org/wiki/Verdier_duality</a> it says the following: Let $F$ be a field, and $X$ a finite dimensional (dimension is defined here cohomologically, but for our purposes a finite dimensional manifold will do) locally compact space.</p> <p>In the part about Poincare duality, it says: $H^k(X,F)=[F,X[k]]$. What is the interpretation of this notation? As I see it, $[F,X[k]]$ means $Hom(F,X[k])$ in the derived category. But this means that $X$ is seen as a complex. How? And why would $Hom(F,X[k])$ equal $H^k(X,F)$?</p> http://mathoverflow.net/questions/68647/understanding-the-wiki-page-on-verdier-duality/68653#68653 Answer by DamienC for Understanding (the wiki page on) Verdier duality DamienC 2011-06-23T20:59:48Z 2011-07-07T09:16:07Z <p>I think here is how one should understand the last paragraph of the wiki. </p> <p>Consider $f:X\to pt$. We have (all functors are derived and my $Hom$ are sheaf $Hom$) $$Hom_{pt}(f_!F,F)=f_*Hom_X(F,f^!F)$$</p> <p>$f_!F=\Gamma_c(X,F)$, so the l.h.s. computes the dual of the cohomology with compact support of the constant sheaf $F$, i.e. the dual of the cohomology with compact support of $X$. </p> <p>$Hom_X(F,f^!F)= \Gamma(f^!F)$ and thus the r.h.s. is $f_*(\Gamma(f^!F))=\Gamma(X,f^!F)$, the homology of $X$. </p> <p>EDIT : to answer precisely the question I include a summary of the comments. </p> <ol> <li><p>there is a typo in the wiki: $[F,X[k]]$ should be understood as $[F,F[k]]$.</p></li> <li><p>for any sheaf of $F$-modules $S$ (concentrated in degree 0), $[F[−k],S]=H^k(X,S)$. </p></li> <li><p>Contrary to what is claimed in the wiki, there is no duality between $H^k(X,F)=[F[−k],F]$ and $H_k(X,F)=[F[−k],D_X]$ (where $D_X:=f^!F$ for $f:X\to pt$). The duality is, either between $H^k_c(X,F)$ and $H_k(X,F)$, or between $H^k(X,F)$ and $H_k^{BM}(X,F)$. And as far as I understand $H^*(X,S)$ is dual to $H_c^*(X,S^\vee)$. </p></li> </ol> http://mathoverflow.net/questions/68647/understanding-the-wiki-page-on-verdier-duality/94478#94478 Answer by Justin Curry for Understanding (the wiki page on) Verdier duality Justin Curry 2012-04-19T00:39:00Z 2012-04-19T00:39:00Z <p>I just revamped what was written. Perhaps now it is more understandable:<a href="http://en.wikipedia.org/wiki/Verdier_duality" rel="nofollow">New Wiki Entry on Verdier duality.</a></p>